SOLUTION: the third term of a geometric sequence is 4 and the sixth term is 32/27. find the nth term.

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Question 1127066: the third term of a geometric sequence is 4 and the sixth term is 32/27. find the nth term.
Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!
the third term of a geometric sequence is 4
a%5B3%5D=4
and the sixth term is 32%2F27
a%5B6%5D=32%2F27
find the nth term:
The formula for the general term for each geometric sequence is a%5Bn%5D=a%5B1%5D%2Ar%5E%28n-1%29 where a%5B1%5D is first term and r is common ratio
we need to find a%5B1%5D and r

since given a%5B3%5D=4, we know that n=3 then

a%5B3%5D=a%5B1%5D%2Ar%5E%28n-1%29
4=a%5B1%5D%2Ar%5E%283-1%29
4=a%5B1%5D%2Ar%5E2....solve for a%5B1%5D
a%5B1%5D+=4%2Fr%5E2........eq.1

since given a%5B6%5D=32%2F27, we know that n=6 then
a%5B6%5D=a%5B1%5D%2Ar%5E%286-1%29
32%2F27=a%5B1%5D%2Ar%5E5.....solve for a%5B1%5D
a%5B1%5D+=%2832%2F27%29%2Fr%5E5
a%5B1%5D+=32%2F%2827r%5E5%29.......eq.2

from eq.1 and eq.2 we have

32%2F%2827r%5E5%29=4%2Fr%5E2....solve for r
32r%5E2=4%2A27r%5E5
cross%2832%298%2Ar%5E2=cross%284%2927r%5E5
8r%5E2=27r%5E5
r%5E5%2Fr%5E2=8%2F27
r%5E3=8%2F27
r%5E3=2%5E3%2F3%5E3
r%5E3=%282%2F3%29%5E3
r=%282%2F3%29

now find a%5B1%5D
a%5B1%5D+=4%2Fr%5E2........eq.1
a%5B1%5D+=4%2F%282%2F3%29%5E2
a%5B1%5D+=4%2F%284%2F9%29
a%5B1%5D+=%289%2A4%29%2F4
a%5B1%5D+=9

the nth term formula is: a%5Bn%5D=9%2A%282%2F3%29%5E%28n-1%29